Change of Variable Formula in Lebesgue integral

Question

I'm studying the change of variable formula in Lebesgue integral, and I've found some similar materials in here.

Among the explanation, I'm not sure why $F \circ g$ is differentiable a.e

If it meant that $F \circ g$ is absolutely continuous, I guess there is some counter-example that makes the composition not absolutely continuous such as this

What property makes the composition differentiable a.e? Any comments about this including my error (if exists) would be grateful. Thank you.

-additional-

I have additional questions regarding the answer.

1: I suppose when $f$ is real-valued function (actually I'm interested in this general case), I could define $f_n$ as follows and then apply rest of your methods (including DCT). Tell me if this is wrong. $$f_n(x) = \begin{cases}n &\text{ if } f(x) \ge n\\ -n &\text{ if } f(x) \le -n\\ f(x) &\text{ otherwise}\end{cases} $$

2: I think the prerequisite for this question is estimated to be the following statement; $f$ is integrable, and $g$ is absolutely continuous and increasing, then $f(g(x))g'(x)$ is integrable as well. This was my trial, which seems not perfect.

Suppose $g(x) = y$, then this leads to $g'(x)dx = dy$ (basic calculus). Then,

$\int |f(g(x))g'(x)|dx$

=$\int |f(y)|dy$ < $\infty$ (∵ $f$ is integrable)

But this doesn't look nice to show the integrability of $f(g(x))g'(x)$. Is there any other way to show $f(g(x))g'(x)$ is integrable? Or, is this enough to prove the integrability?

Answer 1

Using notations from the question you originally cite, define the functions $F : g([a,b])\to\mathbb R$ and $G:[a,b]\to\mathbb R$ by

$$F(x) = \int_{g(a)}^x f(t)\ dt\ \text{ and }\ G(x)=\int_a^x f(g(t))g'(t)\ dt $$

Where $g:[a,b]\to\mathbb R$ is a strictly increasing absolutely continuous function and $f:[c,d]\supseteq g([a,b])\to\mathbb R$ is non-negative and integrable.

Now for any integer $n\ge 1$, denote by $f_n$ the truncation of $f$ to level $n$, i.e. $$f_n(x) := f(x) \wedge n = \begin{cases}n &\text{ if } f(x)\ge n\\ f(x) &\text{ otherwise}\end{cases} $$

And accordingly denote by $F_n$ and $G_n$ the integrals $F_n(x) := \int_{g(a)}^x f_n(t)\ dt $ and $G_n(x) := \int_a^x f_n(g(t))g'(t)\ dt$.

Now for any $n\ge 1$ and $y, x\in g([a,b])$, we have

$$\begin{align*}|F_n(y)-F_n(x)|&=\left|\int_x^y f_n(t)\ dt\right|\\ &\leq\int_x^y|f_n(t)|\ dt\\ &\leq\int_x^y n\ dt\le n|y-x| \end{align*} $$

Hence $F_n$ is Lipschitz continuous and therefore $F_n\circ g$ is absolutely continuous, hence a.e. differentiable. Now you can argue as in the previous answer to see that $F_n\circ g (b) = G_n(b)$, that is $$\int_{g(a)}^{g(b)} f_n(t)\ dt = \int_a^b f_n(g(t))g'(t)\ dt $$ Now you can apply the Dominated Convergence Theorem and conclude by letting $n\to\infty$.